31.2 Solutions to Differential Equations: An Introduction 1001 19. Let M = M(t) be the amount of money in a trust fund earning interest at an annual interest rate of r compounded continuously. Suppose money is withdrawn at a rate of w dollars per year. Assuming the withdrawals are being made continuously, we can use a differential equation to model the situation. Analysis of this differential equation shows that the rate of change of money in the account at t = 0 could be either positive or negative, depending on the size of M(0) = M 0 . Find the threshold value of M 0 . Then, by analyzing the sign of the first and second derivatives, argue that if M 0 is less than this threshold value, then M(t) is decreasing and concave down, and if M 0 is greater than this threshold value, then M(t) is increasing and concave up. 20. A miser spends money at a rate proportional to the amount he has. Suppose that right now he has $100,000 stowed under his mattress; he does not pay any taxes and does not earn any return on his money. Assume that this is all the money he has and that he has no other source of income. At the moment he is spending the money at a rate of $10,000 per year. (a) At what rate will he be spending money when he has $50,000? (b) At what time will the amount of money be down to $10,000? 21. A drosophila colony (a colony of fruit flies) is being kept in a laboratory for study. It is being provided with essentially unlimited resources, so if left to grow, the colony will grow at a rate proportional to its size. If we let N(t) be the number of drosophila in the colony at time t, t given in weeks, then the proportionality constant is k. (a) Write a differential equation reflecting the situation. (b) Solve the differential equation using N 0 to represent N(0). (c) Suppose the drosophila are being cultivated to provide a source for genetic study, and therefore drosophila are being siphoned off at a rate of S drosophila per week. Modify the differential equation given in part (a) to reflect the siphoning off. (d) One of your classmates is convinced that the solution to the differential equation in part (c) is given by N(t) = N 0 e kt − St. Show him that this is not a solution to the differential equation. (e) Your classmate is having a hard time giving up the solution he brought up in part (d). He sees that it does not satisfy the differential equation, but he still has a strong gut feeling that it ought to be right. Convince him that it is wrong by using a more intuitive argument. Use words and talk about fruit flies. 22. Solve the differential equations below. Find the general solution. (a) dy dt = sin 3t (b) dy dt = 5 · 2 t (c) dx dt = t+1 t (d) dx dt = t+1 t 2 23. Solve the differential equations below. Find the general solution. (a) dy dt = 3t + 5 (b) dy dt = 3y (c) dy dt =−y (d) dy dt = 0 (e) dy dt = 3y − 6 1002 CHAPTER 31 Differential Equations 24. For each differential equation below, sketch the slope field and find the general solution. (a) dy dt =−y (b) dy dt =−t (c) dy dt =e −t 25. Each function below is a solution to one of the second order differential equations listed. To each function match the appropriate differential equation. C 1 and C 2 are constants. Differential Equations I. d 2 x dt 2 −9x =0 II. d 2 x dt 2 +9x =0 III. d 2 x dt 2 =3x Solution Functions (a) x(t) =5e 3t (b) x(t) =−2e √ 3t (c) x(t) =7 sin 3t (d) x(t) =C 1 sin 3t + C 2 cos 3t (e) x(t) =C 1 e √ 3t + C 2 e − √ 3t 26. For what value(s) of β,ifany,is (a) y = C 1 sin βt a solution to y = 16y? (b) y = C 2 cos βt a solution to y = 16y? (c) y = C 3 e βt a solution to y = 16y? 27. For what value(s) of β,ifany,is (a) y = C 1 sin βt a solution to y =−16y? (b) y = C 2 cos βt a solution to y =−16y? (c) y = C 3 e βt a solution to y =−16y? 28. (a) There are two values of λ such that y = e λ t is a solution to y + 7y + 12y = 0. Find them and label them λ, and λ 2 . (b) Let y = C 1 e λ 1 t + C 2 e λ 2 t , where C 1 and C 2 are arbitrary constants. Verify that y(t) is a solution to y + 7y + 12y = 0. 29. (a) Find λ such that y = e λt is a solution to y + 4y + 4y = 0. (b) Verify that y = te λt is also a solution to y + 4y + 4y + 0. 31.3 QUALITATIVE ANALYSIS OF SOLUTIONS TO AUTONOMOUS DIFFERENTIAL EQUATIONS One way of getting information about the behavior of solutions to a differential equation is to use the differential equation itself to sketch a picture of the solution curves. This will not give us a formula for the solutions; however, a graphical approach will often give us enough qualitative information about the solutions to answer some important questions. In this section we will focus exclusively on autonomous differential equations, differ- ential equations of the form dy dt =f(y). ◆ EXAMPLE 31.14 Newton’s law of cooling, revisited. Suppose a hot or cold beverage is put in a room that is kept at 65 degrees. Then the rate of change of the temperature of the beverage is proportional 31.3 Qualitative Analysis of dy dt = f(y) 1003 to the difference between the temperature of the room and the temperature of the drink. dT dt = k(65 − T), where k is a positive constant and T = T(t)is the temperature of the beverage at time t. Answer the following questions by taking a graphical perspective. (a) What must the temperature of the beverage be in order for its temperature to remain constant? (b) For what temperatures is the beverage cooling down? In other words, where is dT dt negative? (c) Sketch representative solution curves corresponding to a variety of initial conditions. SOLUTION Solving a differential equation involving dT dt means finding temperature, T , as a function of time, t, so we label the vertical axis T and the horizontal axis t. (a) If a quantity is not changing, then its derivative must be zero. dT dt = k(65 − T)is zero only if T = 65. The beverage’s initial temperature must be 65 ◦ in order to remain constant. This agrees perfectly with our intuition. (b) Because the equation dT dt = k(65 − T)is a continuous function of T , the sign of dT dt can only change around the zeros of dT dt ; dT dt = 0 only at T = 65. We draw the T number line vertically because temperature, T , is plotted on the vertical axis. When T>65, dT dt is negative; this makes sense because a hot beverage will cool off. When T<65, dT dt is positive; this makes sense because a cold beverage will warm up. (c) We use the information contained in Figure 31.3. dT dt = k(65 − T),sothe further T is from 65 the greater the magnitude of the slope and the closer T is to 65 the more gentle the slope. 6565 T(t) is decreasing T(t) is increasing dT dt < 0 dT dt = 0 dT dt > 0 TT t Figure 31.3 1004 CHAPTER 31 Differential Equations 65 65° T t Figure 31.4 The nonconstant solutions are asymptoticto the constant solution, T = 65, meaning that the temperature of the drink will approach the temperature of the room. The constant solution T(t)=65 is called the equilibrium solution. ◆ Definition An equilibrium solution to a differential equation is a solution that is constant for all values of the independent variable (often t = time). If dy dt = f(y),then the equilibrium solutions can be found by setting dy dt = 0 and solving for y. Equilibrium solutions are also referred to as constant solutions. The next example will serve as a case study of differential equations of the form dx dt = f(y), where f(y)is continuous. ◆ EXAMPLE 31.15 Do a qualitative analysis of the solutions to the differential equation dy dt = (y − 1)(y − 3). Sketch representatives of the family of solutions. SOLUTION Solving the differential equation means finding y as a function of t, so we label the vertical axis y and the horizontal axis t. First we identify the equilibrium, or constant solutions, solutions for which dy dt = 0. These correspond to horizontal lines in the ty-plane. The constant solutions are y = 1 and y = 3. Because dy dt is continuous, the sign of dy dt can only change around the zeros of dy dt . For y>3, dy dt > 0 ⇒ y(t) is increasing. For 1 <y<3, dy dt < 0 ⇒ y(t) is decreasing. For y<1, dy dt > 0 ⇒ y(t) is increasing. 31.3 Qualitative Analysis of dy dt = f(y) 1005 y t y = 3 y = 1 3 1 2 y t y = 3 y = 1 3 1 2 y 3 1 (+) (–) (+) sign of dy dt graph of y Figure 31.5 The equilibrium solutions of dy dt = f(y)divide the plane into horizontal strips. f(y)is continuous so two distinct solutions to the differential equation cannot intersect. Therefore, each nonconstant solution lies completely within one strip. We check the sign of dy dt in each of these strips. Because f(y) is a continuous function, the sign of dy dt within each strip does not change. Therefore each nonconstant solution is either strictly increasing or strictly decreasing. Representative solution curves are given in Figure 31.6. Notethat within each horizontal band the solutions are of the same form. (In each band one arbitrary solution has been highlighted; the other solutions within the band can be obtained by shifting the highlighted solution horizontally.) y B A t 3 1 2 Figure 31.6 Each solution in a bounded strip is asymptotic to a constant solution. (Why? The solutions are horizontal translates and there will not be an “empty” horizontal band.) Each solution in an unbounded strip is either asymptotic to a constant solution or increases or decreases without bound. ◆ In summary, given a differential equation of the form dy dt = f(y), where f(y) is continuous we know the following: The solution curves will not intersect. The constant solutions (equilibrium solutions) partition the ty-plane into horizontal strips in which each solution is of the same type (i.e., the solutions are horizontal translates of one another). Within each horizontal strip the solution curve is either strictly increasing or strictly decreasing because the sign of dy dt can only change around the zeros of f(y). 1006 CHAPTER 31 Differential Equations Every solution curve is either asymptotic to a constant solution or increases or decreases without bound. Equilibria and Stability Equilibrium solutions to differential equations can be classified as stable, unstable, or semistable. In Example 31.5, y = 1 and y = 3 are the equilibrium solutions. The equilibrium at y = 1 is referred to as a stable equilibrium; under slight perturbation the system will tend back toward the equilibrium. For instance, if y is slightly less than 1, as t →∞,y increases toward 1. Similarly, if y is a bit more than 1 (i.e., anything under 3), then as t →∞,ydecreases toward 1. The equilibrium at y = 3isanunstable equilibrium; under slight perturbation the system does not return to this equilibrium. If y is slightly greater than 3, then as t →∞,y increases without bound. On the other hand, if y is slightly less than 3, then as t →∞,ytends away from 3 and toward the stable equilibrium of 1. As illustration, consider two stationary coins, one lying flat and the other balancing on its edge. Since their positions are not changing with time, both coins are at equilibrium. The former configuration is stable under slight perturbations. The latter, however, is unstable because the coin balancing on its edge will topple under small perturbations. In terms of modeling real-world situations, stable equilibrium solutions are states that systems would naturally gravitate toward, while unstable equilibrium solutions are thresholds between two qualitatively different types of outcomes. In any modeling situation knowing whether an equilibrium solution is stable or unstable is of tremendous importance, because real life is chock full of perturbations. y y = c semistable y = b stable y = a y y = c y = b y = a unstable Figure 31.7 ◆ EXAMPLE 31.16 Find and classify the equilibrium solutions of dx dt = x 2 − x. SOLUTION To find the equilibrium solutions, set dx dt = 0. dx dt = x(x − 1) = 0 x = 0orx=1 Now look at the sign of dx dt on either side of the equilibrium values. 31.3 Qualitative Analysis of dy dt = f(y) 1007 x t x = 1 x t x = 11 0 x (+) (+) (–) sign of dx dt Figure 31.8 x(t) = 0 is astable equilibrium solution, while x(t) = 1 is an unstable equilibrium solution. ◆ EXERCISE 31.4 Do a qualitative analysis of the solutions of the differential equation dS dt = 0.01S(300 − S). (This is the differential equation that arose in modeling the spread of a flu in a college dormitory.) Sketch some representative solution curves and any constant solutions. Classify the equilibria. After analyzing the solutions in the abstract, determine what this says about the spread of the flu. EXERCISE 31.5 Write a differential equation whose solution curves look like those sketched below. (Begin by constructing a differential equation with equilibrium solutions at y = 3 and y =−2.) y t y = 3 y = –2–2 3 Figure 31.9 ◆ EXAMPLE 31.17 An industrial plant produces radioactive material at a constant rate of 4 kilograms per year. The radioactive material decays at a rate proportional to the amount present and has a half- life of 20 years. (a) Write a differential equation whose solution is R(t), the amount of material present t years after this practice begins. (b) Sketch some representative solutions corresponding to different initial values of R. Include the equilibrium solution. Can we predict the level of radioactive material in the long run? SOLUTION (a) The rate of production is constant at 4 kilograms per year regardless of the time and regardless of the amount present. 1008 CHAPTER 31 Differential Equations The rate out is proportional to the amount present, so it is kR, where k is a constant. rate of change = rate of production − rate of decay dR dt = 4 − kR We must find the value of k using the half-life information. We know that if a substance decays at a rate proportional to the amount present (and there are no further additions to the amount), then it can be modeled by dS dt = kS. The solution to this differential equation is S(t) = Ce kt . We know the half-life to be 20 years, so we can compute k. S(20) = 1 2 Ce 0t Ce 20k = 0.5C e 20k = 0.5 20k = ln 0.5 k = − ln 2 20 ≈−0.0347 In our case, since dR dt = 4− kR already incorporates the negative sign indicating decay, k = ln 2 20 ≈ 0.0347. The differential equation is dR dt = 4 − ln 2 20 R. (b) The equilibrium solution occurs where dR dt = 0. 4 = ln 2 20 R ⇒ R = 80 ln 2 ≈ 115.42 kg. This is a stable equilibrium. R(t) R(t) = t 80 ln 2 Figure 31.10 In the long run, the amount of radioactive material present approaches 80 ln 2 kg, or approximately 115.42 kg. ◆ Logistic Population Growth Because unlimited resources are not observed in the real world, population growth is not exponential indefinitely. At a certain point members of the population begin to compete with one another for limited resources and the growth rate slows. The larger the population, the more prominent the role of competition. Ecologists and population biologists call the number of animals a particular environment can support the carrying capacity of that environment for that animal. Suppose a Tanzanian savannah has the carrying capacity L for lions, where L is a fixed constant. Let P = P(t) be the population of lions at time t. We expect the graph of P versus t to look something like the graph drawn in Figure 31.11 below. 31.3 Qualitative Analysis of dy dt = f(y) 1009 P P = L t Figure 31.11 For P small the graph should be concave up . As P gets closer to the carrying capacity we expect the population growth to slow down and hence the graph of P to be concave down. The most basic observation is that there are constant solutions at P = 0 and P = L;if there are no lions we don’t expect lions to be spontaneously generated, and if the population is at the carrying capacity we expect it to stay there (barring natural or unnatural disasters). To formulate a mathematical model for the situation we’ll write a differential equation that has equilibrium solutions at P = 0 and P = L and is increasing for P ∈ (0, L). dP dt = kP(L − P) will work, provided k is a positive constant. NOTE This equation, dP dt = kLP − kP 2 , can be thought of as dP dt = k 1 P − k 2 P 2 , where k 1 and k 2 are constant. Without the −k 2 P 2 term this would just be the familiar exponential growth differential equation. But because lions are competing with one another for limited resources we need to incorporate a braking factor. This braking factor cannot be of the form kP, as this would still give exponential growth; only the proportionality constant would have changed. A braking factor of the form k(P · P) is reasonable because the competition is proportional to “interactions” between lions and lions. A population growth model of the form dP dt = kP(L − P) is known as a logistic growth model. The logistic model was first used in the early 1800s by the Belgian demographer Pierre-Franc¸ois Verthust to model human world population. The population figures he predicted for 100 years into the future were off by less than 1%. ◆ EXAMPLE 31.18 Suppose the number of fish in a lake grows according to the equation dP dt = 0.45P − 0.0005P 2 . (a) What is the lake’s carrying capacity for fish? Is it a stable equilibrium? (b) What size is the fish population when it is growing most rapidly? 1010 CHAPTER 31 Differential Equations SOLUTION (a) Let’s sketch representative solution curves. We begin by identifying the equilibrium solutions. dP dt = P(0.45 − 0.0005P)= 0 P = 0 or 0.0005P = 0.45 P = 0.45 0.0005 = 900 PP t stable 900900 0 P t 900 (+) (–) sign of dp dt Figure 31.12 From this analysis we see that the carrying capacity is 900 fish and that this is a stable equilibrium. (b) The fish population is growing most rapidly at the point of inflection of the curves lying in the strip between P = 0 and P = 900. Our analysis so far tells us that there ought to be at least one such point for each curve because the curve is asymptotic to both P = 0 and P = 900. We will show that for each solution curve in this interval there is only one such point of inflection and determine for what value of P this occurs. There are two different approaches we can take to this problem. Approach 1. We must determine the value of P in the interval (0, 900) such that dP dt is maximum. We want to know where 0.45P − 0.0005P 2 is maximum. But the expression 0.45P − 0.0005P 2 is quadratic, corresponding to a parabola with P -intercepts of 0 and 900, a parabola opening downward; hence its maximum value is at P = −0.45 2(−0.0005) = 0.45 0.01 = 450. The point of inflection is at P = 450. Approach 2. dP dt will be maximum at the point at which the solution curve changes concavity. The point of inflection of P(t)is the point at which the sign of d 2 P dt 2 changes. dP dt = 0.45P − 0.0005P 2 Differentiate both sides of the equation above with respect to t keeping in mind that P itself is a function of t. . asymptotic to a constant solution. (Why? The solutions are horizontal translates and there will not be an “empty” horizontal band.) Each solution in an unbounded strip is either asymptotic to a. population biologists call the number of animals a particular environment can support the carrying capacity of that environment for that animal. Suppose a Tanzanian savannah has the carrying capacity. 31.4 The nonconstant solutions are asymptoticto the constant solution, T = 65, meaning that the temperature of the drink will approach the temperature of the room. The constant solution T(t)=65